有效地从 Scala 中的字符串中提取日期

Question

我想从 Scala 中具有以下形式的多个字符串中提取日期（例如 2015-01-01）：

val s = "basedir/somedir/tmp/BLAH/2015-01-01.txt"

我知道我可以通过基本的字符串拆分-trim-剥离操作来实现这一点，但是在 Scala 中是否有更简洁的方法来实现它？我可以使用 Scala 提供的一些不错的正则表达式 "hidden features" 来做到这一点吗？

我试过了，但没有成功：

val s = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
val regex = "(\d+)-(\d+)-(\d+).txt"
val regex(year, month, date) = s

Answer 1

使用正则表达式提取器进行模式匹配

val regex = ".*/(\d{4}-\d{2}-\d{2}).txt".r //remove / after .* if you think its not needed.

str match {
  case regex(date) => Some(date)
  case _ => None
}

使用上面的代码而不是下面的代码，因为下面的代码会在运行时导致匹配错误。

val regex(a) = "basedir/somedir/tmp/BLAH/2015-01-01.txt"

而不是正则表达式前面的 .*。您可以使用 unanchored.

val regex = "(\d{4}-\d{2}-\d{2}).txt".r.unanchored

Scala REPL

scala>  val regex = "(\d{4}-\d{2}-\d{2}).txt".r.unanchored
regex: scala.util.matching.UnanchoredRegex = (\d{4}-\d{2}-\d{2}).txt

scala> val regex(a) = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
a: String = 2015-01-01

Scala REPL

scala> val regex = ".*/(\d{4}-\d{2}-\d{2}).txt".r
regex: scala.util.matching.Regex = .*/(\d{4}-\d{2}-\d{2}).txt

scala> val regex(a) = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
a: String = 2015-01-01

Scala REPL

scala> val str = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
str: String = basedir/somedir/tmp/BLAH/2015-01-01.txt

scala>  val regex = ".*/(\d{4}-\d{2}-\d{2}).txt".r
regex: scala.util.matching.Regex = .*/(\d{4}-\d{2}-\d{2}).txt

scala>
     |     str match {
     |       case regex(date) => Some(date)
     |       case _ => None
     |     }
res21: Option[String] = Some(2015-01-01)

如果你也想匹配目录那么

scala val regex = ".*/(.*)/(\d{4}-\d{2}-\d{2}).txt".r
regex: scala.util.matching.Regex = .*/(.*)/(\d{4}-\d{2}-\d{2}).txt

scala> val s = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
s: String = "basedir/somedir/tmp/BLAH/2015-01-01.txt"

scala> val regex(dir, date) = s
dir: String = "BLAH"
date: String = "2015-01-01"

Answer 2

昨天我被谴责解析一些日志文件，每行有三四个不同的时间表示。

我推荐最少的正则表达式和最多类型的东西。

$ scala
Welcome to Scala 2.12.0 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_101).
Type in expressions for evaluation. Or try :help.

scala> val s = "basedir/somedir/tmp/BLAH/2015-01-01.txt"
s: String = basedir/somedir/tmp/BLAH/2015-01-01.txt

scala> val r = raw".*/([\d-]*)\.txt".r
r: scala.util.matching.Regex = .*/([\d-]*)\.txt

scala> val r(date) = s
date: String = 2015-01-01

scala> import java.time._, format._, DateTimeFormatter._
import java.time._
import format._
import DateTimeFormatter._

scala> ISO_LOCAL_DATE.parse(date)
res0: java.time.temporal.TemporalAccessor = {},ISO resolved to 2015-01-01

scala> Instant.from(res0)
java.time.DateTimeException: Unable to obtain Instant from TemporalAccessor: {},ISO resolved to 2015-01-01 of type java.time.format.Parsed
  at java.time.Instant.from(Instant.java:378)
  ... 27 elided
Caused by: java.time.temporal.UnsupportedTemporalTypeException: Unsupported field: InstantSeconds
  at java.time.format.Parsed.getLong(Parsed.java:203)
  at java.time.Instant.from(Instant.java:373)
  ... 27 more

scala> LocalDate.from(res0)
res2: java.time.LocalDate = 2015-01-01

为方便起见的其他变体：

scala> object LocalDateX { def unapply(s: String): Option[LocalDate] = util.Try(LocalDate.from(ISO_LOCAL_DATE.parse(s))).toOption }
defined object LocalDateX

scala> val r(LocalDateX(date)) = s
date: java.time.LocalDate = 2015-01-01

或

scala> implicit class RContext(val sc: StringContext) {
     | object r {
     |   def apply(args: Any*) = sc.s(args: _*)
     |   def unapplySeq(s: String) = sc.parts.mkString("(.+)").r.unapplySeq(s)
     | }}
defined class RContext

scala> val r".*/${LocalDateX(date)}.txt" = s
date: java.time.LocalDate = 2015-01-01

Answer 3

您可以分两行执行此操作。

import java.text.SimpleDateFormat
val date_format = new java.text.SimpleDateFormat("yyyy-MM-dd")
date_format.format(date_format.parse("2017-10-26 09:15:54.127"))
res39: String = 2017-10-26

有效地从 Scala 中的字符串中提取日期

extract date from string in Scala efficiently

java

scala

scala-collections

scalatest